OP Malhotra Class 12 Maths Solutions Chapter 26 Application of Calculus in Commerce and Economics Exercise 26 (C)

Question 1. The demand for a certain product is represented by the equation \( p = 20 + 5x - 3x^2 \) where units demanded and;? is the price per unit. Find
(i) Total revenue
(ii) The marginal revenue
(iii) The marginal revenue when 2 units are sold.
Answer:
Given the demand function: \( p = 20 + 5x - 3x^2 \)
(i) To find the total revenue function \( R(x) \), we multiply the price per unit (p) by the number of units sold (x).
\( R(x) = p \times x \)
\( R(x) = (20 + 5x - 3x^2)x \)
\( R(x) = 20x + 5x^2 - 3x^3 \)
(ii) To find the marginal revenue function (MR), we differentiate the total revenue function \( R(x) \) with respect to x. This shows the change in total revenue from selling one additional unit.
\( MR = \frac{d}{dx} R(x) \)
\( MR = \frac{d}{dx} (20x + 5x^2 - 3x^3) \)
\( MR = 20 + 10x - 9x^2 \)
(iii) To find the marginal revenue when 2 units are sold, we substitute \( x=2 \) into the MR function.
\( MR_{x=2} = 20 + 10(2) - 9(2^2) \)
\( MR_{x=2} = 20 + 20 - 9(4) \)
\( MR_{x=2} = 40 - 36 \)
\( MR_{x=2} = 4 \)
In simple words: First, we find the total money from sales by multiplying price by quantity. Then, we find the extra money we get from selling one more item by using a mathematical tool called a derivative. Finally, we put the number 2 into this extra-money formula to find the specific marginal revenue for 2 units.

🎯 Exam Tip: Remember that total revenue is \( p \times x \), and marginal revenue is the first derivative of the total revenue function. Pay close attention to the differentiation rules for polynomials and the correct substitution of x values.

Question 2.
(i) The demand function of a monopolist is given by \( p = 100 - x - x^2 \). Find (a) the revenue function, (b) marginal revenue function.
(ii) If \( p=\frac{150}{q-5}-6 q \) represents the demand function for a product, where p is the price per unit for q units. Determine the marginal revenue function.
(iii) If \( p=\frac{5}{q-5}-6 q \) represents the demand function for a product, where p is the price per unit of q units, find the marginal revenue.
Answer:
(i) Given demand function: \( p = 100 - x - x^2 \)
(a) Revenue function \( R(x) \):
To find the total revenue, multiply the price (p) by the quantity (x).
\( R(x) = p \times x \)
\( R(x) = (100 - x - x^2)x \)
\( R(x) = 100x - x^2 - x^3 \)
(b) Marginal revenue function (MR):
To find the marginal revenue, differentiate the total revenue function \( R(x) \) with respect to x.
\( MR = \frac{d}{dx} R(x) \)
\( MR = \frac{d}{dx} (100x - x^2 - x^3) \)
\( MR = 100 - 2x - 3x^2 \)
(ii) Given demand function for this part (following the solution steps): \( p = \frac{150}{q^2+2} - 4 \)
Revenue function \( R(q) \):
\( R(q) = p \times q \)
\( R(q) = \left( \frac{150}{q^2+2} - 4 \right) q \)
\( R(q) = \frac{150q}{q^2+2} - 4q \)
Marginal revenue function (MR):
To find the marginal revenue, differentiate the total revenue function \( R(q) \) with respect to q. This tells us the rate at which total revenue changes as one more unit is sold.
\( MR = \frac{d}{dq} R(q) \)
\( MR = \frac{d}{dq} \left( \frac{150q}{q^2+2} - 4q \right) \)
\( MR = 150 \frac{d}{dq} \left( \frac{q}{q^2+2} \right) - \frac{d}{dq} (4q) \)
Using the quotient rule \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \):
\( MR = 150 \left( \frac{1 \cdot (q^2+2) - q \cdot (2q)}{(q^2+2)^2} \right) - 4 \)
\( MR = 150 \left( \frac{q^2+2 - 2q^2}{(q^2+2)^2} \right) - 4 \)
\( MR = 150 \left( \frac{2 - q^2}{(q^2+2)^2} \right) - 4 \)
(iii) Given demand function: \( p = \frac{5}{q-5} - 6q \)
Revenue function \( R(q) \):
\( R(q) = p \times q \)
\( R(q) = \left( \frac{5}{q-5} - 6q \right) q \)
\( R(q) = \frac{5q}{q-5} - 6q^2 \)
Marginal revenue function (MR):
To find the marginal revenue, differentiate \( R(q) \) with respect to q.
\( MR = \frac{d}{dq} R(q) \)
\( MR = \frac{d}{dq} \left( \frac{5q}{q-5} - 6q^2 \right) \)
\( MR = 5 \frac{d}{dq} \left( \frac{q}{q-5} \right) - \frac{d}{dq} (6q^2) \)
Using the quotient rule:
\( MR = 5 \left( \frac{1 \cdot (q-5) - q \cdot (1)}{(q-5)^2} \right) - 12q \)
\( MR = 5 \left( \frac{q-5 - q}{(q-5)^2} \right) - 12q \)
\( MR = 5 \left( \frac{-5}{(q-5)^2} \right) - 12q \)
\( MR = \frac{-25}{(q-5)^2} - 12q \)
In simple words: For each part, first we find the total money earned from sales by multiplying the given price equation by the quantity. Then, we find the extra money earned when one more unit is sold by using calculus (differentiation). This shows how the total revenue changes with quantity.

🎯 Exam Tip: Be careful with the quotient rule for differentiation, especially when dealing with complex fractions. Make sure to apply it correctly for each term in the revenue function, and double-check your signs and arithmetic.

Question 3. The total revinue received from the sale of x units of a product is given by \( R(x) = 36x + 3x^2 + 5 \). Find.
(i) the average revenue
(ii) the marginal revenue
(iii) the marginaXand average revenue when x = 5
(iv) the actual revenue from selling 50 th item.
Answer:
Given total revenue function: \( R(x) = 36x + 3x^2 + 5 \)
(i) To find the average revenue function (AR), divide the total revinue \( R(x) \) by x.
\( AR = \frac{R(x)}{x} \)
\( AR = \frac{36x + 3x^2 + 5}{x} \)
\( AR = 36 + 3x + \frac{5}{x} \)
(ii) To find the marginal revenue function (MR), differentiate the total revinue function \( R(x) \) with respect to x. This tells us the additional revenue from selling one more unit.
\( MR = \frac{d}{dx} R(x) \)
\( MR = \frac{d}{dx} (36x + 3x^2 + 5) \)
\( MR = 36 + 6x \)
(iii) To find the average revenue and marginal revenue when \( x = 5 \):
Average Revenue at \( x=5 \):
\( AR_{x=5} = 36 + 3(5) + \frac{5}{5} \)
\( AR_{x=5} = 36 + 15 + 1 \)
\( AR_{x=5} = 52 \)
Marginal Revenue at \( x=5 \):
\( MR_{x=5} = 36 + 6(5) \)
\( MR_{x=5} = 36 + 30 \)
\( MR_{x=5} = 66 \)
(iv) To find the actual revenue from selling the 50th item, calculate the difference between the total revenue from 50 items and 49 items.
Required actual revenue from selling 50th item = \( R(50) - R(49) \)
\( R(50) - R(49) = (36 \times 50 + 3 \times 50^2 + 5) - (36 \times 49 + 3 \times 49^2 + 5) \)
\( R(50) - R(49) = 36(50 - 49) + 3(50^2 - 49^2) \)
\( R(50) - R(49) = 36(1) + 3(50 - 49)(50 + 49) \)
\( R(50) - R(49) = 36 + 3(1)(99) \)
\( R(50) - R(49) = 36 + 297 \)
\( R(50) - R(49) = 333 \)
This calculation provides the incremental revenue from the last item sold, offering insight into unit-specific profitability.
In simple words: We start with the total money earned from sales. Then we divide by quantity to get average money per item, and use a derivative to get the extra money from one more item. We then use these formulas for when 5 items are sold. For the 50th item, we find the extra money it brings by subtracting the total money from 49 items from the total money from 50 items.

🎯 Exam Tip: Remember that "actual revenue from selling the nth item" is found by \( R(n) - R(n-1) \), not just \( MR(n) \). This is a common point of confusion. Always read the question carefully to determine if marginal revenue or actual revenue for a specific unit is required.

Question 4. The demand function for a monopolist is given by \( x = 100 - 4p \). Find
(i) total revenue function
(ii) average revenue function
(iii) marginal revenue function
(iv) price and quantity at which MR =0
Answer:
Given the demand function: \( x = 100 - 4p \)
First, we rearrange this equation to express price (p) in terms of quantity (x):
\( 4p = 100 - x \)
\( \implies \) \( p = \frac{100 - x}{4} \)
\( p = 25 - \frac{x}{4} \)
(i) To find the total revenue function \( R(x) \), multiply price (p) by quantity (x).
\( R(x) = p \times x \)
\( R(x) = \left( 25 - \frac{x}{4} \right) x \)
\( R(x) = 25x - \frac{x^2}{4} \)
(ii) To find the average revenue function (AR), divide the total revenue \( R(x) \) by x.
\( AR = \frac{R(x)}{x} \)
\( AR = \frac{25x - \frac{x^2}{4}}{x} \)
\( AR = 25 - \frac{x}{4} \)
(iii) To find the marginal revenue function (MR), differentiate the total revenue function \( R(x) \) with respect to x. This shows the additional revenue generated by selling one more unit.
\( MR = \frac{d}{dx} R(x) \)
\( MR = \frac{d}{dx} \left( 25x - \frac{x^2}{4} \right) \)
\( MR = 25 - \frac{2x}{4} \)
\( MR = 25 - \frac{x}{2} \)
(iv) To find the price and quantity when marginal revenue (MR) is 0, set MR to 0 and solve for x.
When \( MR = 0 \):
\( 25 - \frac{x}{2} = 0 \)
\( \implies \) \( \frac{x}{2} = 25 \)
\( \implies \) \( x = 50 \text{ units} \)
Now, substitute \( x = 50 \) into the price function \( p = 25 - \frac{x}{4} \) to find the required price.
\( p = 25 - \frac{50}{4} \)
\( p = 25 - 12.5 \)
\( p = \text{Rs } 12.5 \)
At this price and quantity, selling an additional unit would not increase total revenue.
In simple words: First, change the demand equation to show price based on quantity. Then, find total money from sales by multiplying price by quantity. After that, calculate average money per item by dividing total money by quantity, and extra money from one more item by using a derivative. Finally, set this extra-money value to zero to find the quantity and then the price.

🎯 Exam Tip: Always make sure to express price (p) as a function of quantity (x) before calculating revenue functions. This is a crucial first step for many calculus problems in economics, as revenue is typically \( R(x) = p(x) \times x \).

Question 5. A monopolist demand function for one of its products is \( p(x) = ax + b \). He knows that he can sell 1400 units when the price is Rs 4 per unit and he can sell 1800 units at a price of Rs 2 per unit. Find the total average and marginal revenue functions. Also find the price per unit when the marginal revenue is zero.
Answer:
Given the demand function: \( p(x) = ax + b \) ...(1)
We are given two points (quantity, price):
When \( x = 1400 \), \( p = \text{Rs } 4 \):
Substituting these values into equation (1): \( 4 = 1400a + b \) ...(2)
When \( x = 1800 \), \( p = \text{Rs } 2 \):
Substituting these values into equation (1): \( 2 = 1800a + b \) ...(3)
Now, subtract equation (2) from equation (3) to find the value of 'a':
\( (2 - 4) = (1800a + b) - (1400a + b) \)
\( -2 = 400a \)
\( \implies \) \( a = \frac{-2}{400} \)
\( a = \frac{-1}{200} \)
Substitute the value of \( a = \frac{-1}{200} \) into equation (2) to find the value of 'b':
\( 4 = 1400 \left( \frac{-1}{200} \right) + b \)
\( 4 = -7 + b \)
\( \implies \) \( b = 11 \)
So, the demand function is (substituting 'a' and 'b' into (1)):
\( p(x) = -\frac{1}{200}x + 11 \)
Or, \( p(x) = 11 - \frac{x}{200} \)

Next, we find the revenue functions based on this demand function:
**Total Revenue Function \( R(x) \):**
This is found by multiplying price (p) by quantity (x).
\( R(x) = p(x) \times x \)
\( R(x) = \left( 11 - \frac{x}{200} \right) x \)
\( R(x) = 11x - \frac{x^2}{200} \)

**Average Revenue Function (AR):**
This is found by dividing total revenue \( R(x) \) by quantity (x).
\( AR = \frac{R(x)}{x} \)
\( AR = \frac{11x - \frac{x^2}{200}}{x} \)
\( AR = 11 - \frac{x}{200} \)

**Marginal Revenue Function (MR):**
This is found by differentiating total revenue \( R(x) \) with respect to x.
\( MR = \frac{d}{dx} R(x) \)
\( MR = \frac{d}{dx} \left( 11x - \frac{x^2}{200} \right) \)
\( MR = 11 - \frac{2x}{200} \)
\( MR = 11 - \frac{x}{100} \)

Finally, find the price and quantity when marginal revenue (MR) is zero:
Set \( MR = 0 \):
\( 11 - \frac{x}{100} = 0 \)
\( \implies \) \( \frac{x}{100} = 11 \)
\( \implies \) \( x = 1100 \text{ units} \)
Substitute \( x = 1100 \) into the demand function to find the corresponding price:
\( p = 11 - \frac{1100}{200} \)
\( p = 11 - 5.5 \)
\( p = \text{Rs } 5.50 \)
At this specific price and quantity, selling an additional unit would not contribute any extra revenue.
In simple words: First, we use the given price and quantity details to figure out the exact demand equation. Then, we use this equation to calculate the total money earned, the average money made per item, and the extra money from selling one more item. Lastly, we find the price and quantity where selling another item would bring in no additional money.

🎯 Exam Tip: When given multiple price-quantity points, use them to set up a system of linear equations to find the 'a' and 'b' coefficients for the demand function first. This is a common and critical initial step for such problems.

Question 6. Suppose the consumers will demand 40 units of a product when the price is Rs 12 per unit and 25 units when the price is Rs 18 each. Find the demand function, assuming that is linear. Also, determine the total revenue function, the average revenue function and the marginal revenue function.
Answer:
Let the linear demand function be \( x = ap + b \) ...(1), where x is the number of units demanded and p is the price per unit.
We are given two points (quantity, price):
When \( x = 40 \), \( p = \text{Rs } 12 \):
Substitute these values into equation (1): \( 40 = 12a + b \) ...(2)
When \( x = 25 \), \( p = \text{Rs } 18 \):
Substitute these values into equation (1): \( 25 = 18a + b \) ...(3)
Subtract equation (2) from equation (3) to find the value of 'a':
\( (25 - 40) = (18a + b) - (12a + b) \)
\( -15 = 6a \)
\( \implies \) \( a = \frac{-15}{6} \)
\( a = \frac{-5}{2} \)
Substitute the value of \( a = \frac{-5}{2} \) into equation (3) to find the value of 'b':
\( 25 = 18 \left( \frac{-5}{2} \right) + b \)
\( 25 = 9 \times (-5) + b \)
\( 25 = -45 + b \)
\( \implies \) \( b = 70 \)
So, the demand function is (substituting 'a' and 'b' into (1)):
\( x = -\frac{5}{2}p + 70 \)
Now, to prepare for revenue calculations, we express p in terms of x:
\( \frac{5}{2}p = 70 - x \)
\( \implies \) \( p = \frac{2}{5}(70 - x) \)
\( p = \frac{140 - 2x}{5} \)
\( p = 28 - \frac{2}{5}x \)

Next, we determine the revenue functions based on this price function:
**Total Revenue Function \( R(x) \):**
\( R(x) = p \times x \)
\( R(x) = \left( 28 - \frac{2}{5}x \right) x \)
\( R(x) = 28x - \frac{2}{5}x^2 \)

**Average Revenue Function (AR):**
\( AR = \frac{R(x)}{x} \)
\( AR = \frac{28x - \frac{2}{5}x^2}{x} \)
\( AR = 28 - \frac{2}{5}x \)

**Marginal Revenue Function (MR):**
\( MR = \frac{d}{dx} R(x) \)
\( MR = \frac{d}{dx} \left( 28x - \frac{2}{5}x^2 \right) \)
\( MR = 28 - \frac{2}{5} (2x) \)
\( MR = 28 - \frac{4}{5}x \)
These functions describe how the revenue changes with the number of units sold, which is crucial for business decisions.
In simple words: We find the demand rule by using the given price and quantity pairs to make a linear equation. Then, using this rule, we calculate the total money earned, the average money per item, and the extra money from selling one more item by using derivatives.

🎯 Exam Tip: When forming the linear demand function, decide whether to express x in terms of p or p in terms of x first, based on which makes the subsequent revenue calculations simpler. Often, \( p = f(x) \) is more convenient for calculating revenue functions.

Question 7. A firm knows that the demand function for one of its products is linear. It also knows that it can sell 1000 units when the price is Rs 4 per unit, and it can sell 1500 units when the price is Rs 2 a unit. Determine
(i) the demand function,
(ii) the total revenue function,
(iii) the average revenue function,
(iv) the marginal revenue function.
Answer:
(i) Let the linear demand function be \( x = ap + b \), where x is the number of units demanded and p is the price per unit.
We are given two points (quantity, price):
When \( x = 1000 \), \( p = \text{Rs } 4 \):
Substitute these values into the demand function: \( 1000 = 4a + b \) ...(1)
When \( x = 1500 \), \( p = \text{Rs } 2 \):
Substitute these values into the demand function: \( 1500 = 2a + b \) ...(2)
Subtract equation (1) from equation (2) to find the value of 'a':
\( (1500 - 1000) = (2a + b) - (4a + b) \)
\( 500 = -2a \)
\( \implies \) \( a = -250 \)
Substitute the value of \( a = -250 \) into equation (1) to find the value of 'b':
\( 1000 = 4(-250) + b \)
\( 1000 = -1000 + b \)
\( \implies \) \( b = 2000 \)
So, the demand function is: \( x = -250p + 2000 \)
To prepare for revenue calculations, we express p in terms of x:
\( 250p = 2000 - x \)
\( \implies \) \( p = \frac{2000 - x}{250} \)
\( p = 8 - \frac{x}{250} \)

(ii) To find the total revenue function \( R(x) \), multiply price (p) by quantity (x).
\( R(x) = p \times x \)
\( R(x) = \left( 8 - \frac{x}{250} \right) x \)
\( R(x) = 8x - \frac{x^2}{250} \)

(iii) To find the average revenue function (AR), divide the total revenue \( R(x) \) by x.
\( AR = \frac{R(x)}{x} \)
\( AR = \frac{8x - \frac{x^2}{250}}{x} \)
\( AR = 8 - \frac{x}{250} \)

(iv) To find the marginal revenue function (MR), differentiate the total revenue function \( R(x) \) with respect to x.
\( MR = \frac{d}{dx} R(x) \)
\( MR = \frac{d}{dx} \left( 8x - \frac{x^2}{250} \right) \)
\( MR = 8 - \frac{2x}{250} \)
\( MR = 8 - \frac{x}{125} \)
The average revenue function in this case is identical to the price function, which is a common characteristic when price is expressed as a function of quantity.
In simple words: First, we use the given sales data to find the demand equation that relates price and quantity. Then, using this equation, we calculate the total money earned, the average money per item sold, and the extra money received from selling one more item by using derivatives.

🎯 Exam Tip: When given x and p points, always clearly define your demand function \( (x = ap + b \text{ or } p = ax + b) \) at the start. Ensure consistency throughout your calculations for 'a' and 'b' to avoid errors.

Question 8. Find the relationship between the slopes of marginal revenue curve and the average revenue curve, for the demand function \( p = a - bx \).
Answer:
Given the demand function: \( p = a - bx \)

First, find the Total Revenue function \( R(x) \):
\( R(x) = p \times x \)
\( R(x) = (a - bx)x \)
\( R(x) = ax - bx^2 \)

Next, find the Average Revenue function (AR):
This is the total revenue divided by the quantity.
\( AR = \frac{R(x)}{x} \)
\( AR = \frac{ax - bx^2}{x} \)
\( AR = a - bx \)

Then, find the slope of the Average Revenue (AR) curve by differentiating AR with respect to x:
Slope of AR curve = \( \frac{d}{dx} (AR) \)
Slope of AR curve = \( \frac{d}{dx} (a - bx) \)
Slope of AR curve = \( -b \)

Now, find the Marginal Revenue function (MR):
This is the derivative of the total revenue function with respect to x.
\( MR = \frac{d}{dx} R(x) \)
\( MR = \frac{d}{dx} (ax - bx^2) \)
\( MR = a - 2bx \)

Finally, find the slope of the Marginal Revenue (MR) curve by differentiating MR with respect to x:
Slope of MR curve = \( \frac{d}{dx} (MR) \)
Slope of MR curve = \( \frac{d}{dx} (a - 2bx) \)
Slope of MR curve = \( -2b \)

Comparing the slopes:
The slope of the MR curve (\( -2b \)) is twice the slope of the AR curve (\( -b \)). This fundamental relationship holds true for any linear demand function in economics.
In simple words: We first find the formulas for how much money is earned per item on average and how much extra money is earned from selling one more item. Then we take the derivative of each formula to find how steeply their lines go up or down. We notice that the steepness of the extra-money line is always twice the steepness of the average-money line.

🎯 Exam Tip: For any linear demand curve, the marginal revenue curve will always have twice the slope of the average revenue curve. Understanding this relationship can help you quickly verify your calculations in related problems.

Question 9. In the production unit of a firm it is found that the total number of unites produced is dependent upon the number of workers and is obtained by the relation \( x = 25n(n^3 + 36)^{-1/2} \). The demand function of the product is \( p = \frac{250}{x+15} \). Determine the marginal revenue when \( n = 4 \).
Answer:
Given the demand function: \( p = \frac{250}{x+15} \)

First, find the Total Revenue function \( R(x) \):
This is calculated by multiplying price (p) by quantity (x).
\( R(x) = p \times x \)
\( R(x) = \left( \frac{250}{x+15} \right) x \)
\( R(x) = \frac{250x}{x+15} \)

Next, find the Marginal Revenue function (MR) by differentiating \( R(x) \) with respect to x. We use the quotient rule, where \( u = 250x \) and \( v = x+15 \).
\( u' = 250 \)
\( v' = 1 \)
\( MR = \frac{d}{dx} R(x) = \frac{u'v - uv'}{v^2} \)
\( MR = \frac{250(x+15) - 250x(1)}{(x+15)^2} \)
\( MR = \frac{250x + (250 \times 15) - 250x}{(x+15)^2} \)
\( MR = \frac{3750}{(x+15)^2} \) ...(1)

Now, we need to find the value of x (total units produced) when the number of workers \( n=4 \), using the given relation between x and n:
\( x = 25n(n^3 + 36)^{-1/2} \)
Substitute \( n = 4 \) into this equation:
\( x = 25(4)(4^3 + 36)^{-1/2} \)
\( x = 100(64 + 36)^{-1/2} \)
\( x = 100(100)^{-1/2} \)
\( x = 100 \left( \frac{1}{\sqrt{100}} \right) \)
\( x = 100 \left( \frac{1}{10} \right) \)
\( x = 10 \text{ units} \)

Finally, substitute \( x = 10 \) into the Marginal Revenue function (1) to find MR when \( n=4 \):
\( MR = \frac{3750}{(10+15)^2} \)
\( MR = \frac{3750}{(25)^2} \)
\( MR = \frac{3750}{625} \)
\( MR = 6 \)
This means that when 4 workers are employed, and 10 units are produced, the marginal revenue from selling an additional unit is Rs 6.
In simple words: First, we find the formula for the total money earned from sales. Then, we use calculus to get the formula for the extra money earned from selling one more item. Next, we figure out how many items are made when 4 workers are on the job. Finally, we put this number of items into the extra-money formula to get our final answer.

🎯 Exam Tip: This question requires sequential calculation: first derive \( R(x) \), then \( MR(x) \), then find x in terms of n, and finally substitute the given n to get x and then MR. Be methodical in each step to ensure accuracy.

Question 10. For the demand function \( p=\frac{a}{x+b}-c \), where \( ab > 0 \), show that the narginal revenue decreases with the increase of x.
Answer:
Given the demand function: \( p = \frac{a}{x+b}-c \)
We are also given that \( ab > 0 \).

First, find the Total Revenue function \( R(x) \):
\( R(x) = p \times x \)
\( R(x) = \left( \frac{a}{x+b}-c \right) x \)
\( R(x) = \frac{ax}{x+b} - cx \)

Next, find the Marginal Revenue function (MR) by differentiating \( R(x) \) with respect to x. This function shows how total revenue changes with each additional unit sold.
\( MR = \frac{d}{dx} R(x) \)
\( MR = \frac{d}{dx} \left( \frac{ax}{x+b} - cx \right) \)
To differentiate \( \frac{ax}{x+b} \), we use the quotient rule: \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \), where \( u=ax \) and \( v=x+b \).
\( u' = a \)
\( v' = 1 \)
So, \( \frac{d}{dx} \left( \frac{ax}{x+b} \right) = \frac{a(x+b) - ax(1)}{(x+b)^2} = \frac{ax+ab-ax}{(x+b)^2} = \frac{ab}{(x+b)^2} \)
And \( \frac{d}{dx} (cx) = c \)
Therefore, the Marginal Revenue function is:
\( MR = \frac{ab}{(x+b)^2} - c \)

To show that marginal revenue decreases with an increase in x, we need to find the derivative of MR with respect to x. If this second derivative is negative, it means MR is a decreasing function.
\( \frac{d}{dx} (MR) = \frac{d}{dx} \left( \frac{ab}{(x+b)^2} - c \right) \)
\( \frac{d}{dx} (MR) = \frac{d}{dx} (ab(x+b)^{-2} - c) \)
\( \frac{d}{dx} (MR) = ab(-2)(x+b)^{-3}(1) - 0 \)
\( \frac{d}{dx} (MR) = \frac{-2ab}{(x+b)^3} \)

We are given that \( ab > 0 \). For practical applications, \( x > 0 \) (number of units). Assuming \( b \) is positive or such that \( (x+b) \) is positive (e.g., if \( b \) is negative, \( x \) is large enough), then \( (x+b)^3 \) will be positive.
Since \( ab > 0 \), the term \( -2ab \) will be a negative number.
A negative number divided by a positive number results in a negative number.
Therefore, \( \frac{-2ab}{(x+b)^3} < 0 \) for all \( x > 0 \) and \( ab > 0 \).
Since the derivative of the marginal revenue function is negative, it means that the marginal revenue decreases as x increases. This implies that as more units are produced and sold, the additional revenue from each extra unit becomes smaller, a common economic principle known as diminishing marginal returns.
In simple words: We first find the formula for the total money from sales, then we find the extra money gained from selling one more item by using a derivative. To prove that this extra money goes down as more items are sold, we take another derivative of the extra-money formula. Since the problem tells us that 'ab' is a positive number, the result of this second derivative is negative, which confirms that the extra money earned decreases as sales increase.

🎯 Exam Tip: To prove a function is decreasing, always calculate its first derivative. If this derivative is consistently negative under the given conditions, the function is proven to be decreasing. Pay attention to all given conditions, such as \( ab > 0 \), as they are crucial for the final conclusion.